If n is the positive integer and if 1-4x-4x2n-r-02narxr- where ar are real numbers- The value of a2n-1 is

The correct option is A n2^2n ( 1+4x+4 x ^ 2 ) #160;^ n =( 1+2x ) #160;^ 2n Value of coefficient of 2n 1 is #160; a _ 2n 1 =_ #16 ...

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Properties of Composite Function

If n is the...

Question

If $n$ is the positive integer and if $(1 + 4 x + 4 x^{2})^{n} = 2 n \sum r = 0 a_{r} x^{r}$ , where $a_{r}$ are real numbers. The value of $a_{2 n - 1}$ is

2^{2 n}

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n - 1 2^{2 n}

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n 2^{2 n}

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(n + 1) 12^{2 n}

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Solution

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Similar questions

n

(1 + 4 x + 4 x^{2})^{n} = 2 n \sum r = 0 a_{r} x^{r}

a_{r}

a_{2 n - 1}

n \in N

{(1 + 4 x + 4 x^{2})}^{n} = 2 n \sum r = 0 a_{r} x^{r}

a_{o}, a_{1}, a_{2}, . . . . . . ., a_{2 n}

a_{2 n - 1}

n \in N

(1 + 4 x + 4 x^{2})^{n} = \sum_{r = 0}^{2 n} a_{r} x^{r}

a_{0}, a_{1}, a_{1}, a_{2}, . . . . . . . . ., a_{2 n}

2 \sum_{r = 0}^{n} a_{2 r}

n

(1 + 4 x + 4 x^{2})^{n} = 2 n \sum r = 0 a_{r} x^{r}

a_{r}

2 n \sum r = 1 a_{2 r - 1}

n \in N

{(1 + 4 x + 4 x^{2})}^{n} = 2 n \sum r = 0 a_{r} X^{r},

a_{o}, a_{1}, a_{2}, . . . . . . ., a_{2 n}

n \sum r = 0 a_{2 r}

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